Re: Peano's Axioms are Inconsistent

On Aug 7, 8:57 pm, Joshua Cranmer <Pidgeo...@xxxxxxxxxxxxxxx> wrote:
RussellE wrote:
Which property of first order logic is it?
It seems like you are just assuming such a y exists.

You completely missed the part where I showed that substituting n + 1
for y satisfies the predicate, and then the discussion that n + 1 must
exist (since x is also n + 1, and we know that x exists because of the
Ax quantifier).

For all x, x is greater than y? That obviously fails for all y, since 1
is not greater than any other natural number. I'm not really sure what
you're tying to say here.

I create a family of sets.
After each step, y, I want the set defined by Ax x>y.

Your notation is horrendous. What you want, I think, is {x : x > y},
which makes a lot more sense than a predicate Ax x>y.

Since BNTM writes to every position in X,
there must exist a step when the set of
blank positions is empty. Every position
means all positions.

Here is where you fall down. Your first statement is the proposition:
AxEy T(y)(x) = '1', which we agree to be true.

The predicate for cell z being blank at step x:
Ay <= x T(z)(y) != '1' (in other words, BNTM never wrote to cell z in a
step before the present).
[Note that we introduce an ordering relation here, one that will be
defined in terms of the + 1 or successor function we've been using.]

Your statement that the set of blank positions being empty is equivalent
to saying that there exists a step x such that no cell z is empty:
Ex Az not (Ay <= x T(z)(y) != '1'), or
Ex Az Ey <= x T(z)(y) = '1'.

To disprove this proposition, I look at the negative:
Ax Ez Ay <= x T(z)(y) != '1'.

This is a more difficult proposition to prove. As a lemma, let me point
out that T(a)(b) = '1' if and only if a = b; this I present without
proof, as it immediately follows from earlier discussion of the
definition of T and the axiom of induction.

T(z)(y) != '1' therefore fails only if y = z. If z > x, then it becomes
a y which is irrelevant to the qualifier, so Ay <= x T(z)(y) = '1' is
true if z > x.

So we reduce the negative to:
Ax Ez z > x, or, for all elements of our set, there exists a larger
element. The few axioms we have been using are insufficient to prove or
disprove this statement.

This negative is equivalent to saying that there exists no largest
element. For the set of natural numbers, this statement is true, which
means that we have formally proved your original proposition false by
proving its negative to be true.

If you want to convince me of a contradiction, you must formally prove
the proposition true.

What the end result means is that, although every cell is written to,
there exists no time at which all cells have been written to--you need
to actually pass infinity to be able to do that.

We agree there is no step when the
set of blank positions is empty.

Then you say BNTM writes to every position.
You can't have it both ways.
You are saying there is a step when the set
of blank positions is empty.
If the set of blank positions is empty, I can
prove there is a step it becomes empty.

Time has nothing to do with it.
You have to prove there exists a step for every
position. And the only way you can do that
is by proving the set of blank positions is empty.

By proving BNTM writes to every position,
you are proving there are no more positions
to write to.

One of the elements of X has no successor.


Russell
- 2 many 2 count
.

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